Problem 4, October 1995Author(s): Diane ThomasSource: The Mathematics Teacher, Vol. 89, No. 3 (March 1996), pp. 255, 263Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27969727 .

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linearly. Your figure 2 is a proof without words.

Foster responds: The solution Kaiman presented for problem 27 is certainly another good solution to the problem. Some students may find it more challenging than the one we presented.

Problem 28 may be solved the way Kaiman suggested or the way we presented it. Students see things in different ways, and each method is appropriate.

Problem 4, October 1995

A baseless cylinder may be formed from an 8.5" 11" sheet in two ways. They have the same lateral surface area. Are their volumes also equal?

A nice general relationship exists between the volumes of two cylin ders each formed by the same

rectangular piece of paper. If 0<a < bf then

6 forms cylinder!

living volume

*\2nj a 4

forme cylinder

having volume

So the two volumes are related as the ratio of the two sides of the original rectangle: Volume of cylinder II

a2b 4

rab^ A

= - ? (Volume

of cylinder i)

(Continued on page 263)

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Vol. 89, No. 3 ? March 1996 255

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EM

Note also that the taller rectangle always produces a smaller volume.

Diane Thomas 6015 Brown Road Oxford, OH 45056

Kennedy responds: A nice obser vation! Of course those volumes are related as the sides of the rectangle?why did I not think of that solution?

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Problem 6, October 1995 Readers may be interested in the following alternative solution to the problem for 6 October 1995:

Points A and are the mid points of the sides of a square whose lengths are 8. Find the area of trapezoid ABCD.

Let W, X, Y, and denote, respectively, the upper-right, upper-left, lower-left, and lower

right vertices of the square; let denote the measure of angle WBZ; and let ( ) denote area. Then {ABCD} = {XYZ} - {ABY} -

{BCZ}-{ADX},{XYZ} = 32, {ABY} = 8, {BCZ} = {ADX}=a, sin0=2A/5, and cos0=lA?5. Thus, we need find only a to com

plete the solution. Let r denote CZ. Then, by the

law of sines,

4 sin?

sin(l35?-0)

jl.jl + j_.a /2 V5 /2 V?

= 8 /2 3

Hence,

a-_?4?-?sin 45? 2 3

= 16

3'

Therefore,

\aBCD} = 32-S-^-^ = ̂. 1 J 3 3 3

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Kennedy responds: Your alterna tive solution is fine and would be especially appropriate for stu dents in a precalculus course or a

straight trigonometry course. Thanks for your interest!

Problems 6 and 17, October 1995 Here is an alternative method for finding the area of trapezoid ABCD. This approach does not

require the use of the area for mula for a trapezoid. Instead, we

can determine its area as the dif ference between the area of half the square?the triangular half it resides in?and the three triangles that fill out that half.

The problem: Points A and are the mid

points of the sides of a square whose lengths are 8. Find the area of trapezoid ABCD.

The area of trapezoid ABCD is

equal to the area of triangle FGH minus the sum of the areas of tri

angles ABG, ADH, and BCF. The area of ABG = (l/2)(4)(4) = 8. Notice that ADH is congruent to BCF, so we need only find the area of triangle BCF. Notice also that since C is on the angle bisec tor of ZEFG, altitudes CP and CQ are congruent. Therefore,

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Vol. 89, No. 3 ? March 1996 263

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